Trying to understand the statistics question (hypothesis testing and significant difference) QUESTION;
A researcher, Warr (summarizing Putnam) claims that increases in watching television has led to fewer hours in the day to be active civically. Test that idea by comparing average days of church attendance in a year comparing people who watch less than average hours of television per day and those who watch above average hours of television per day. 
Average Annual Church Attendance By Above or Below Average TV Watching: 
Hours of TV :
\begin{array}{r|r|r}
& < \text{Avg.} & \le \text{Avg.} \\ \hline
\text{Mean} & 22.3 & 20.8 \\
\text{sd} & 25.6 & 25.5 \\
n & 903 & 390
\end{array}
a. Use hypothesis testing procedures to determine if there is a statistically significant difference in average number of days of church attendance across the TV groups. Be certain to interpret the results. Use $\alpha=0.01$, $2$-tail. 
b. Interpret the results of this statistical test. Do these statistical findings support Warr’s argument? 

SOLUTION (attempt):
So since it is stated to be a two-tail and one sample we use the t-test to generate a number, not sure why we use this formula but here we go:
$$\frac{(22.3-20.8)}{\sqrt{\frac{25.6^2}{903} + \frac{25.5^2}{390}}} = \frac{1.5}{1.547} = 0.970.$$
Alright, so we have $0.970$, not sure what this tells us, but I know that $\alpha = 0.01$, and so that is the upper and lower regions, our critical regions that we want there to be some overlap so that we can reject. If the number falls outside of those ranges they we can support the null hypothesis. Here we know that an $\alpha$ of $0.01$ and being a two-tail will equal to $2.575$.
So we have a $t$-test which generates the number $0.970$ and we have an $\alpha$ that's on a two-tail equal to $2.575$, not sure what this means. Could someone please explain the meaning and relationship between these two numbers. I am confused on this whole concept. Are these numbers statistically supporting or not supporting the argument?
Thank you
 A: This is several paragraphs long. I will try to go through the logic of this
t test step by step. I hope each step makes sense to you.
We wonder whether the population means of hours of church attendance $\mu_1$
for people who watch less TV differs from $\mu_2$ for people who watch more TV.
The best estimates of these population means are the sample means
$\bar X_1 = 22.3$ and $\bar X_2 = 20.8.$ Accordingly, the best estimate
of $\mu_1 - \mu_2$ is $\bar X_1 - \bar X_2 = 22.3 - 20.8 = 1.5.$
So if there is a difference it seems we may have $\mu_1 > \mu_2.$
But the sample means are not perfect estimates of the population means and 1.5 is a small number,
so we wonder whether another similar experiment might give us a negative
value instead of this small positive value. 
To assess the variability of 1.5 we look at the variability of the sample data. The variance of $\bar X_1$ is estimated by $25.6^2/903$ and the variance of $\bar X_2$ is estimated by $25.5^2/390$. Thus the variance of $\bar X_1 - \bar X_2$ is estimated by $25.6^2/903 + 25.5^2/390 = 2.393,$ and the standard
deviation of the difference in sample mean is $\sqrt{2.393} = 1.547.$
Roughly speaking, the margin of error for $1.5$ is about $2(1.547) \approx 3,$
so we can't be sure that 1.5 is reliably a positive number.
More precisely, we look at the t statistic $T = 0.970.$ If $|T| > 2.576$ we say
that the numerator $1.5$ is significantly different from $0,$ and hence
that $\mu_1$ is significantly different from $\mu_2.$ If we did this study
again with another group of TV watchers and church goers (from the same general population), we can't be sure
whether we would see $\bar X_1 < \bar X_2,\,$ $\bar X_1 > \bar X_2$ or
$\bar X_1 \approx \bar X_2.$
The method using the t statistic is more precise because it takes into
account that there is extra uncertainty when both means and standard
deviations have to be estimated.
Even so, we can't claim that TV watching has nothing at all to do with hours spent in church,
but we also can't say that TV watching is associated with church attendance.

Below is a plot of the density function of the t statistic for this problem.
The red vertical broken lines are the 'critical values' at $\pm 2.576.$
The total area beneath the curve is $1;$ the area between the red lines is $99\%.$
The blue line shows the value of the t statistic for your data. In order
to reject the null hypothesis and claim that TV watching is associated with hours in
church, the blue line would have to be outside the red lines (not between them).

